Question

将并行算法分配给前缀和问题有一些问题。我使用openMP进行并行实现。我的代码如下c所示。

结果显示：

seqsum[6] = 28 != parallelsum[6] = 34

请指教。感谢。

#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#include "omp.h"
#include <string.h>
#define N 10 //33554432 // 2 ^ 25
#define NUM_THREADS 4 

void computeparallelprefix(int *iplist, int *_pprefixsum, unsigned long size)
{
  int nthr, *z, *x = _pprefixsum;
  int i, j, tid, work, lo, hi;
#pragma omp parallel shared(nthr,x,z) private(i,j,tid,work,lo,hi)
  {
    int prev_sum;
    memcpy((void *)x, (void *)iplist, sizeof(int)*size);

    // Assume nthr = 2^k
#pragma omp single
    {
      nthr = omp_get_num_threads();
      z = malloc(sizeof(int)*nthr);
    }   
    tid = omp_get_thread_num();
    work = size /nthr + (i = tid < size%nthr ? 1 : 0);
    lo = (size/nthr)*tid + (i==1 ? tid : size%nthr);
    hi = lo + work;
    if (hi > size)
      hi = size;

    // local prefix sum over x
    for(i=lo+1; i<hi; i++)
      x[i] += x[i-1];    

    // local prefix sum for tid
    z[tid] = x[hi-1];
#pragma omp barrier


    // global prefix sum over z
    for(j=1; j<nthr; j=2*j) {
      if (tid >= j)
        z[tid] = z[tid] + z[tid-j];
#pragma omp barrier
    } 

    // Update local prefix sum x
    prev_sum = z[tid] - x[hi-1];
    for(i=lo;i<hi;i++)
      x[i] += prev_sum;
  }
  free(z);
}

void initlist(int *iplist, unsigned long size)
{
  int i;
  for ( i = 0; i < size; i++)
    iplist[i] = i+1;
  //     iplist[i] = rand() % 13;
}

void printlist(int *list, unsigned long size)
{
  int i;
  for(i = 0; i < size; i++) {
    printf("%d ", list[i]);
  }
  printf("\n");
}

void computeseqprefixsum(int *iplist, int *seqprefixsum, unsigned long size)
{
  int i;
  seqprefixsum[0] = iplist[0];
  for(i = 1; i < size; i++) {
    seqprefixsum[i] = seqprefixsum[i-1] + iplist[i];
  }
}

void checkresults(int *seqsum, int *parallelsum, unsigned long size)
{
  int i;
  for(i = 0; i < size; i++)
  {
    if(seqsum[i] != parallelsum[i]) {
      printf("seqsum[%d] = %d != parallelsum[%d] = %d\n", i, seqsum[i], i,
          parallelsum[i]);
      exit(1);
    }
  }
}

int main(int argc, char *argv[])
{
  // seed the rand generator
  srand(time(NULL));
  double seqstart, seqend, parstart, parend, seqtime, partime;

  // initialize list

  int *iplist, *seqprefixsum, *pprefixsum ;

  iplist = (int*) malloc(sizeof(int) * N);
  seqprefixsum = (int*) malloc(sizeof(int) * N);
  pprefixsum = (int*) malloc(sizeof(int) * N);

  if(iplist == NULL || seqprefixsum == NULL || pprefixsum == NULL) {
    printf("memory cannot be allocated\n");
    exit(1);
  }

  initlist(iplist, N);

  seqstart = omp_get_wtime();

  computeseqprefixsum(iplist, seqprefixsum, N);

  seqend = omp_get_wtime();

  seqtime = seqend - seqstart;

  omp_set_num_threads(NUM_THREADS);

  parstart = omp_get_wtime();

  computeparallelprefix(iplist, pprefixsum, N);

  parend= omp_get_wtime();

  partime = parend - parstart;

  checkresults(seqprefixsum, pprefixsum, N);

  printf("Seq Time : %f, Par Time : %f, Speedup : %f\n", seqtime, partime,
      seqtime/partime);

  free(iplist); free(seqprefixsum); free(pprefixsum);

  return 0;
}

Answer 1

您对代码的前缀和有正确的想法。

我不确定为什么你没有得到正确的结果，但我清理了你的代码，我的版本得到了正确的结果。有关详细信息parallel-cumulative-prefix-sums-in-openmp-communicating-values-between-thread

，请参阅以下问题

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include "omp.h"
#define N 11 //33554432 // 2 ^ 25

void computeparallelprefix(int *iplist, int *_pprefixsum, unsigned long size)
{
  int nthr, *z, *x = _pprefixsum;
  #pragma omp parallel
  {
    int i;
    #pragma omp single
    {
      nthr = omp_get_num_threads();
      z = malloc(sizeof(int)*nthr+1);
      z[0] = 0;
    }
    int tid = omp_get_thread_num();
    int sum = 0;
    #pragma omp for schedule(static) 
    for(i=0; i<size; i++) {
      sum += iplist[i];
      x[i] = sum;
    }
    z[tid+1] = sum;
    #pragma omp barrier

    int offset = 0;
    for(i=0; i<(tid+1); i++) {
        offset += z[i];
    }

    #pragma omp for schedule(static)
    for(i=0; i<size; i++) {
      x[i] += offset;
    }
  }
  free(z);
}

int main(void ) {

  int *iplist, *pprefixsum ;

  iplist = (int*) malloc(sizeof(int) * N);
  pprefixsum = (int*) malloc(sizeof(int) * N);

  for(int i=0; i<N; i++) iplist[i] = i+1;
  for(int i=0; i<N; i++) printf("%d ", iplist[i]); printf("\n");

  computeparallelprefix(iplist, pprefixsum, N);
  for(int i=0; i<N; i++) printf("%d ", pprefixsum[i]); printf("\n");
  for(int i=0; i<N; i++) printf("%d ", (i+1)*(i+2)/2); printf("\n");
  return 0;
}

使用openMP（c代码）并行计算前缀和

1 个答案: